Virtual laboratory/Physics/Relativity/Relativistic velocity addition

Relativistic velocity addition

Add two velocities close to the speed of light. The classical v₁+v₂ could exceed c, but the relativistic formula u=(v₁+v₂)/(1+v₁v₂/c²) always gives a result smaller than c. Compare the two approaches on a velocity axis.

Relativistic velocity addition
Physics · Theory of relativity · Velocity addition (FIZ.REL.PP.02)
Velocity v1v_1 (in units of c)0.70 c
Velocity v2v_2 (in units of c)0.70 c
Adding two velocities close to c, the result:
velocity v₁
0.70 c
velocity v₂
0.70 c
classical sum
1.400 c
relativistic result u
0.940 c
🎯 Challenge: hit the result u
streak: 0

I'll give you a target resultant velocity — tune v₁ and v₂ to reach it. u = (v₁+v₂)/(1+v₁v₂/c²), always u < c.

What just happened?

You added two velocities close to c. “Common sense” (the classical sum) gives a result greater than c, but nature does not allow it — the relativistic formula pulls the result back below the barrier c.

Relativistic velocity addition:

u=v1+v21+v1v2c2u = \dfrac{v_1 + v_2}{1 + \dfrac{v_1 v_2}{c^2}}

— for small velocities the denominator ≈ 1 and we recover the ordinary u≈v1+v2u \approx v_1 + v_2. As the velocities grow, the denominator grows and “trims” the sum so that always u<cu < c.

What does it mean in practice?

Take v1=v2=0.9 cv_1 = v_2 = 0.9\,c. Classically it would be 1.8 c — impossible. Relativistically:

u=0.9c+0.9c1+0.81=1.8c1.81≈0.994 cu = \dfrac{0.9c + 0.9c}{1 + 0.81} = \dfrac{1.8c}{1.81} \approx 0.994\,c

Why can’t light be “caught up with”?

Substitute v2=cv_2 = c: u=(v1+c)/(1+v1/c)=cu = (v_1+c)/(1+v_1/c) = c. Adding anything to c again gives c — the speed of light is the same in every reference frame.

Why you need this for the exam:

Problems about “rockets sending a signal” and questions “can c be exceeded” are solved by a single formula. Trick: compute the denominator 1+v1v2/c21 + v_1 v_2/c^2 and check that the result always comes out smaller than c.

Remember: velocities do not add “simply” — the relativistic formula makes the result never exceed c. The speed of light is an insurmountable barrier.

Illustrative model · NexTutor Virtual laboratory

Enjoyed the experiment? Share it.

It's free — send it to friends from school.

Theory & formulas

Classical formula

With ordinary (Galilean) velocity addition: ucl=v1+v2u_{cl} = v_1 + v_2 For velocities close to cc this would give a result exceeding the speed of light — which is impossible.

Relativistic formula

The correct addition of velocities along a single line: u=v1+v21+v1v2c2u = \dfrac{v_1 + v_2}{1 + \dfrac{v_1 v_2}{c^2}} The result always satisfies u<cu < c, even when both velocities are close to cc.

When the classical formula is enough

For small velocities (v≪cv \ll c) the term v1v2/c2v_1 v_2 / c^2 is negligible and u≈v1+v2u \approx v_1 + v_2. The speed of light cc is the limiting velocity.

Notation: v1, v2v_1,\ v_2 — the velocities being added, uu — the resultant (relativistic) velocity, uclu_{cl} — the classical result, cc — the speed of light.

Frequently asked questions

›

Can the speed of light be exceeded by adding velocities?

No — the relativistic formula u=(v₁+v₂)/(1+v₁v₂/c²) always gives u<c, even when both velocities are close to c.

›

When is the classical formula enough?

For small velocities (v≪c), when the term v₁v₂/c² is negligible and u≈v₁+v₂.

›

What is the limiting velocity?

The speed of light c — no body or signal can exceed it.

Teaching? Embed this experiment on your site.

Free widget — paste it on your school site, blog or notes. It links to the full version.

Teaching a class or tutoring? Guide students in NexTutor →

Related experiments

Mass–energy equivalence →The Michelson–Morley experiment — the end of the ether →Length contraction (Lorentz contraction) →Time dilation →